Two point charges of +2.0 μC and -6.0 μC are located on the x-axis at x = -1.0 cm and x = +2.0 cm respectively. Where should a third charge of +3.0-μC be placed on the +x-axis so that the potential at the origin is equal to zero? (k = 1/4πε0 = 8.99 × 109 N · m2/C2)

Final answer:

To connect 90-W, 120-V light bulbs to a 20-A, 120-V circuit without tripping the breaker, you can connect a maximum of 26 bulbs.

Explanation:

To determine how many 90-W, 120-V light bulbs can be connected to a 20-A, 120-V circuit without tripping the circuit breaker, we need to calculate the total power consumption of the light bulbs and compare it to the circuit's maximum current capacity.

Using the formula: Power (P) = Voltage (V) x Current (I), we can rearrange it to solve for current: I = P/V.

For each bulb, I = 90 W / 120 V = 0.75 A.

Given the maximum current capacity of the circuit is 20 A, divide the total current capacity by the current per bulb:

Total bulbs = 20 A / 0.75 A = 26.67 bulbs.

Since we can't have a fraction of a bulb, the maximum number of 90-W, 120-V light bulbs that can be connected to the circuit without tripping the breaker is 26 bulbs.

Final answer:

By calculating the current drawn by each 90-W, 120-V light bulb, we find that about 26 bulbs can be connected to a 20-A, 120-V circuit without exceeding the current limit and tripping the circuit breaker.

Explanation:

The question involves calculating the number of 90-W, 120-V light bulbs that can be connected to a 20-A, 120-V circuit. First, it is important to understand that the power rating of a light bulb (like 90 W) indicates the power consumed when it operates at its specified voltage (120 V in this case).

The power, voltage, and current are related by the formula Power (P) = Voltage (V) × Current (I).

Therefore, the current for one 90-W bulb can be found by rearranging the formula:

I = P/V, which gives I = 90 W / 120 V = 0.75 A.

Since the circuit can handle up to 20 A, you divide the total allowable current by the current per bulb: 20 A / 0.75 A/bulb = approximately 26.67.

This means you can connect 26 bulbs without tripping the circuit breaker, since you cannot connect a fraction of a bulb.

Answer:

D

Explanation:

After each half life, half the original amount remains.

So after two half lives, (1/2)^2 remains, or 1/4.

Answer:

Magnetic force, [tex]F=1.12\times 10^{-13}\ N[/tex]

Explanation:

It is given that,

Velocity of proton, [tex]v=1.8\times 10^6\ m/s[/tex]

Angle between velocity and the magnetic field, θ = 53°

Magnetic field, B = 0.49 T

The mass of proton, [tex]m=1.672\times 10^{-27}\ kg[/tex]

The charge on proton, [tex]q=1.6\times 10^{-19}\ C[/tex]

The magnitude of magnetic force is given by :

[tex]F=qvB\ sin\theta[/tex]

[tex]F=1.6\times 10^{-19}\ C\times 1.8\times 10^6\ m/s\times 0.49\ Tsin(53)[/tex]

[tex]F=1.12\times 10^{-13}\ N[/tex]

So, the magnitude of the magnetic force on the proton is [tex]1.12\times 10^{-13}\ N[/tex]. Hence, this is the required solution.

Answer:

Focal length, f = 0.43 meters

Explanation:

It is given that,

Power of the reading glasses, P = 2.3 D

We need to calculate the focal length of the reading glasses. The relationship between the power and the focal length inverse i.e.

[tex]P=\dfrac{1}{f}[/tex]

[tex]f=\dfrac{1}{P}[/tex]

[tex]f=\dfrac{1}{2.3\ D}[/tex]

f = 0.43 m

So, the focal length of the reading glasses found on the rack in a drugstore is 0.43 meters.

The focal length of 2.30 diopters reading glasses is approximately 0.435 meters or 43.5 centimeters.

The focal length of reading glasses, which is the distance over which they focus light, is inversely related to their power in diopters. In this case, the reading glasses have a power of 2.30 diopters (D). To calculate the focal length (f) in meters, use the formula f = 1/P, where P is the power in diopters. Therefore, for glasses with a power of 2.30 D, the focal length would be f = 1/2.30, which equals approximately 0.435m (or 43.5 cm).

Answer:

0.3 m

Explanation:

Initially, the package has both gravitational potential energy and kinetic energy.  The spring has elastic energy.  After the package is brought to rest, all the energy is stored in the spring.

Initial energy = final energy

mgh + ½ mv² + ½ kx₁² = ½ kx₂²

Given:

m = 50 kg

g = 9.8 m/s²

h = 8 sin 20º m

v = 2 m/s

k = 30000 N/m

x₁ = 0.05 m

(50)(9.8)(8 sin 20) + ½ (50)(2)² + ½ (30000)(0.05)² = ½ (30000)x₂²

x₂ ≈ 0.314 m

So the spring is compressed 0.314 m from it's natural length.  However, we're asked to find the additional deformation from the original 50mm.

x₂ − x₁

0.314 m − 0.05 m

0.264 m

Rounding to 1 sig-fig, the spring is compressed an additional 0.3 meters.

We use the principles of Kinetic and Potential Energy to establish an energy conservation equation, balancing the initial kinetic energy of the box against the potential energy stored in the spring at rest. Solving this equation gives the maximum additional deformation of the spring.

From the problem, we know that the mass of the package (m) is 50 kg, the initial speed (v) is 2 m/s, the spring's constant (k) is 30 kN/m or 30000 N/m, and the initial compression of the spring (xi) is 50 mm or 0.05 m. We are tasked to determine the maximum additional deformation (xf) of the spring in bringing the package to rest.

Your problem involves Kinetic Energy and Potential Energy concepts. If we consider the instant when the package begins to hit the spring and the moment when it comes to rest, the energy conservation law can be applied, stating that the initial kinetic energy of the box equals the work done by the spring (which is the potential energy stored in it).

So, the kinetic energy of the package when it begins to hit the spring equals 1/2 * m * v^2, and the potential energy of the spring when the package comes to rest equals 1/2 * k * (xi+xf)^2.

Therefore, equating both expressions and resolving the quadratic equation for xf, provides us with the maximum additional deformation of the spring.

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Explanation and answer:

Given:

6 lb is needed to stretch 4 inches beyond natural length.

Need work done to stretch same string from natural length to 8 inches.

Solution:

string stiffness, K

= Force / stretched distance

= 6 lb / 4 inches

= 1.5 lb/inch

Work done on a string of stiffness K

= (Kx^2)/2 lb-in

= 1.5 lb/in *(8 in)^2)/2

= 48 lb-in.

Final answer:

To find the work done in stretching a spring, one calculates the spring constant using Hooke's Law and then applies the work-energy principle. For a spring stretched 8 inches beyond its natural length, with a spring constant derived as 1.5 lb/in, the work done is 48 lb-in.

Explanation:

To calculate the work done in stretching a spring from its natural length to 8 inches beyond it, we use Hooke's Law and the work-energy principle. Given that a force of 6 lb is required to hold the spring stretched 4 inches beyond its natural length, we first find the spring constant, k, using the formula F = kx, where F is the force applied and x is the displacement from the spring's natural length.

So, k = F / x = 6 lb / 4 in = 1.5 lb/in. The work done, W, in stretching the spring from its natural length to 8 inches beyond can be calculated using the formula W = 1/2 k x², where x is the total displacement from the spring's natural length. In this case, x = 8 in, so W = 1/2 * 1.5 lb/in * (8 in)² = 48 lb-in.

Answer:

Both objects travel the same distance.

(c) is correct option

Explanation:

Given that,

Mass of first object = 4.0 kg

Speed of first object = 2.0 m/s

Mass of second object = 1.0 kg

Speed of second object = 4.0 m/s

We need to calculate the stopping distance

For first particle

Using equation of motion

[tex]v^2=u^2+2as[/tex]

Where, v = final velocity

u = initial velocity

s = distance

Put the value in the equation

[tex]0= u^2-2as_{1}[/tex]

[tex]s_{1}=\dfrac{u^2}{2a}[/tex]....(I)

Using newton law

[tex]a=\dfrac{F}{m}[/tex]

Now, put the value of a in equation (I)

[tex]s_{1}=\dfrac{8}{F}[/tex]

Now, For second object

Using equation of motion

[tex]v^2=u^2+2as[/tex]

Put the value in the equation

[tex]0= u^2-2as_{2}[/tex]

[tex]s_{2}=\dfrac{u^2}{2a}[/tex]....(I)

Using newton law

[tex]F = ma[/tex]

[tex]a=\dfrac{F}{m}[/tex]

Now, put the value of a in equation (I)

[tex]s_{2}=\dfrac{8}{F}[/tex]

Hence, Both objects travel the same distance.

The two objects traveled the same distance which is equal to [tex]\frac{8}{F}[/tex].

The given parameters;

The acceleration of each object is calculated as follows;

F = ma

[tex]a = \frac{F}{m} \\\\a_1 = \frac{F}{4} \\\\a_2 = \frac{F}{1} = F[/tex]

The distance traveled by each object is calculated as follows before coming to rest;

[tex]v^2 = u^2 - 2as\\\\when \ the \ objects \ come \ to \ rest , \ v = 0\\\\0 = u^2 - 2as\\\\2as = u^2\\\\s = \frac{u^2}{2a} \\\\s_1 = \frac{(2)^2}{2(F/4)} = \frac{(2)^2}{F/2} \\\\s_1 = \frac{2(2)^2}{F} \\\\s_1 = \frac{8}{F} \\\\s_2 = \frac{(4)^2}{2F} \\\\s_2 = \frac{8}{F}[/tex]

Thus, we can conclude that the both objects traveled the same distance.

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Answer:

The maximum speed of a particle of the string is 0.34 m/s

Explanation:

The equation for the vertical displacement y of the string is given by y :

[tex]y=1.8\times 10^{-3}cos[\pi(13x-60t)][/tex].............(1)

The general equation of transverse wave is given by :

[tex]y=A(kx-\omega t)[/tex].............(2)

On comparing equation (1) and (2) we get,

[tex]\omega=60\pi[/tex]

Speed of a particle of the string is maximum when displacement is equal to zero. Maximum speed is given by :

[tex]v_{max}=A\omega[/tex]

Where, A = amplitude of wave

[tex]\omega=60\pi[/tex]

So, [tex]v_{max}=1.8\times 10^{-3}\times 60\pi[/tex]

[tex]v_{max}=0.34\ m/s[/tex]

Hence, this is the required solution.

The maximum speed of any particle on the string can be determined by the product of the amplitude and the angular frequency of the wave. In this case, the maximum speed is found to be 0.108 m/s.

The equation given represents the vertical displacement of a transverse wave on a string. We're trying to find the maximum speed of a particle of the string. This refers to the highest speed at which any string particle will be moving as the wave passes by.

From the given wave equation, the particle on the string will be moving in a simple harmonic motion. The maximum speed would be at the equilibrium position with a value equal to the product of the amplitude (A) and the angular frequency (ω).

In your case, the given equation is: y = (1.8x10-3)cos[π(13x - 60t)]. In this equation, the angular frequency (ω) is equal to 60 and the amplitude (A) is equal to 1.8 x 10-3.

Therefore, the maximum speed (vmax) of any particle on the string would be given by: vmax = Aω = (1.8 x 10-3) * 60 = 0.108 m/s

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The distance at which the smaller plane will reach its takeoff speed, we can use the equation of motion:

[tex]\[ v^2 = u^2 + 2as \][/tex]

v is the final velocity (takeoff speed) of the smaller plane (40 m/s)

u is the initial velocity (rest) of the smaller plane (0 m/s)

a is the acceleration of both planes (which is the same)

s is the distance covered by the smaller plane before reaching its takeoff speed (what we want to find)

[tex]\[ s = \frac{{v^2 - u^2}}{{2a}} \][/tex]

[tex]\[ s = \frac{{(40 \, \text{m/s})^2 - (0 \, \text{m/s})^2}}{{2a}} \][/tex]

[tex]\[ a = \frac{{v - u}}{{t}} = \frac{{(80 \, \text{m/s}) - (0 \, \text{m/s})}}{{30 \, \text{s}}} \][/tex]

[tex]\[ s = \frac{{(40 \, \text{m/s})^2}}{{2 \times \left(\frac{{80 \, \text{m/s}}}{30 \, \text{s}}\right)}} \]\[ s = \frac{{1600 \, \text{m}^2/\text{s}^2}}{{\frac{{160 \, \text{m/s}}}{30 \, \text{s}}}} \]\[ s = \frac{{1600 \, \text{m}^2/\text{s}^2 \times 30 \, \text{s}}}{{160 \, \text{m/s}}} \]\[ s = \frac{{48,000 \, \text{m}^2/\text{s}}}{{160 \, \text{m/s}}} \]\[ s = 300 \, \text{m} \][/tex]

Therefore, the smaller plane will reach its takeoff speed after traveling a distance of 300 meters.

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By using the equation of motion v² = u² + 2as, we can calculate the acceleration of the first plane based on its initial velocity, final velocity, and distance covered. Then, using the same acceleration and applying it to the second plane's parameters, we can find the distance it covers before reaching its takeoff speed.

To solve this problem, we'll make use of the equation of motion usually represented as v² = u² + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance covered.

For the first plane, we have v = 80 m/s, u = 0 m/s (because the plane starts from rest), and s = 1200 m. Plugging these into our equation, we can solve for a.

Once we have the acceleration, we can then apply it to the second plane to find the distance it covers (s) before reaching its takeoff speed of 40 m/s. Given a = acceleration calculated above and v = 40 m/s, u = 0 m/s (since the second plane also starts from rest), we can rearrange our equation of motion to solve for s.

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Answer:

V2 = 23.8m/s

Explanation:

as we know That,  

p1/(r*g) + (v1^2)/2*g = p2/(r*g) + (v2^2)/2*g  

p1=300kPa=300000Pa  

p2=200kPa=200000Pa  

r=1200kG/m^3  

v1 = 20 m/s  

250+200 = 166.66+(v2^2)/2  

v2=(450-166.66)*2  

v2^2 =566.68  

v2=23.8m/s

Your answer is : V2 = 23.8m/s

The speed of the flow at this second position is 23.8 m/s.

The given parameters:

The speed of the flow at this second position is calculated by applying Bernoulli's equation as follows;

[tex]P_1 + \frac{1}{2} \rho v_1^2= P_2 + \frac{1}{2} \rho v_2 ^2\\\\(P_1 - P_2) + \frac{1}{2} \rho v_1^2 = \frac{1}{2} \rho v_2 ^2\\\\2(P_1-P_2) + \rho v_1^2 = \rho v_2^2\\\\\frac{2(P_1-P_2) }{\rho} + v_1^2 = v_2^2\\\\\sqrt{\frac{2(P_1-P_2) }{\rho} + v_1^2} = v_2\\\\\sqrt{\frac{2(300,000-200,000) }{1200} + 20^2}= v_2 \\\\\sqrt{566.67} = v_2\\\\23.8\ m/s = \ v_2[/tex]

Thus, the speed of the flow at this second position is 23.8 m/s.

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Answer: [tex]813.13(10)^{-7}F[/tex]

Explanation:

The answer is not among the given options. However, this is a good example of the relation between the energy stored in a capacitor [tex]W[/tex]and its capacitance [tex]C[/tex], which is given by the following equation:

[tex]W=\frac{1}{2}CV^{2}[/tex]   (1)

Where:

[tex]W=1275J[/tex]

[tex]V=5.6kV=5.6(10)^{3}V[/tex] is the voltage

[tex]C[/tex] is the capacitance in Farads, the value we want to find

Isolating [tex]C[/tex] from (1):

[tex]C=\frac{2W}{V^{2}}[/tex]   (2)

[tex]C=\frac{2(1275J)}{(5.6(10)^{3}V)^{2}}[/tex]   (3)

Finally:

[tex]C=0.000081313F=813.13(10)^{-7}F[/tex] This is the capacitance of the cardiac defibrillator

Answer:

Option (D)

Explanation:

The chemical formula for normal water is H2O and the chemical formula for heavy water is D2O.

Where D is deuterium which is the isotope of hydrogen.

There are three isotopes of hydrogen.

1H1 it is called protium.

1H2 it is called deuterium.

1H3 it is called tritium.

Heavy water usually refers to water that contains deuterium.

Answer:

(D) any combustible gas

Explanation:

In ideal diesel cycle the working substance is any combustible gas.

Final answer:

The working substance in the ideal diesel cycle, during the initial intake and compression process, is air. Diesel fuel is injected only after the air has been adiabatically compressed and heated to a sufficient temperature for ignition.

Explanation:

In the ideal diesel cycle, the working substance during the intake stroke is air alone. This air is then compressed adiabatically to a high temperature before diesel fuel is injected. It is important to note that the diesel fuel itself is not part of the initial intake but is rather injected after the air has been compressed, which subsequently ignites due to the high temperature from compression.

The compression stroke significantly raises the air's temperature by compressing it adiabatically from state A to state B. Then during the power stroke, the diesel fuel added to the hot, high pressure air causes ignition and continues the cycle. Therefore, the correct answer to the question is (A) air, as it is the only substance involved in the complete cycle from the intake to the compression phase before fuel injection.

Answer:

0.018 J

Explanation:

The work done to bring the charge from infinity to point P is equal to the change in electric potential energy of the charge - so it is given by

[tex]W = q \Delta V[/tex]

where

[tex]q=3.0 \mu C = 3.0 \cdot 10^{-6} C[/tex] is the magnitude of the charge

[tex]\Delta V = 6.0 kV = 6000 V[/tex] is the potential difference between point P and infinity

Substituting into the equation, we find

[tex]W=(3.0\cdot 10^{-6}C)(6000 V)=0.018 J[/tex]

The work required to move a 3.0 μC charge from infinity to point P with an electric potential of 6.0 kV is 0.018 Joules. This is calculated using the formula W = qV. Substituting the given values into the formula yields the answer.

The work required to move a charge in an electric potential is given by the formula:

W = qV

where W is work, q is the charge, and V is the electric potential. Here, the electric potential V at point P is 6.0 kV (or 6000 V), and the charge q is 3.0 μC (or 3.0 × 10⁻⁶ C).

Substitute the values into the formula:

W = (3.0 × 10⁻⁶ C) × (6000 V)

Which simplifies to:

W = 0.018 J

Therefore, the work required of an external agent to move the 3.0 μC charge from infinity to point P is 0.018 Joules.

Answer: [tex]W_{n}=5.724 (10)^{13})N[/tex]

Explanation:

The weight [tex]W[/tex] of a body or object is given by the following formula:

[tex]W=m.g[/tex] (1)

From here we can find the mass of the body:

[tex]m=\frac{W}{g}=67.346 kg[/tex] (2)

Where [tex]m[/tex] is the mass of the body and [tex]g[/tex] is the acceleration due gravity in an especific place (in this case the earth).

In the case of a neutron star, the weight [tex]W_{n}[/tex] is:

[tex]W_{n}=m.g_{n}[/tex] (3)

Where [tex]g_{n}[/tex] is the acceleration due gravity in the neutron star.

Now, the acceleration due gravity (free-fall acceleration) [tex]g[/tex] of a body is given by the following formula:

[tex]g=\frac{GM}{r^{2}}[/tex] (4)

Where:

[tex]G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex] is the gravitational constant

[tex]M[/tex] the mass of the body (the neutron star in this case)

[tex]r[/tex] is the distance from the center of mass of the body to its surface. Assuming the neutron star is a sphere with a diameter [tex]d=24km[/tex], its radius is [tex]r=\frac{d}{2}=12.5km=12500m [/tex]

Substituting (4) and (2) in (3):

[tex]W_{n}=m(\frac{GM}{r^{2}})[/tex] (5)

[tex]W_{n}=(67.346kg)(\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(1.99(10)^{30}kg)}{(12500m)^{2}})[/tex] (6)

Finally:

[tex]W_{n}=5.724 (10)^{13})N[/tex]

If you weigh 660 N on Earth, you would weigh approximately[tex]\(1.133 \times 10^{-10}\)[/tex]N on the surface of a neutron star with the same mass as our sun and a diameter of 25.0 km.

To find the weight of an object on the surface of a neutron star, you can use the formula for gravitational force:

[tex]\[ F = \frac{{G \cdot M \cdot m}}{{r^2}} \][/tex]

where:

[tex]\( F \)[/tex] is the gravitational force,

[tex]\( G \)[/tex]  is the gravitational constant[tex](\(6.67 \times 10^{-11} \ \text{N} \cdot \text{m}^2/\text{kg}^2\))[/tex],

[tex]\( M \)\\[/tex] is the mass of the neutron star (in this case, the mass of the sun, [tex]\(1.99 \times 10^{30} \ \text{kg}\))[/tex],

[tex]\( m \)[/tex] is the mass of the object (your mass in this case),

[tex]\( r \)[/tex] is the distance from the center of the object to the center of mass of the object (the radius in this case).

The weight (W) of an object is given by[tex]\( W = m \cdot g \)[/tex], where (g ) is the acceleration due to gravity on the object's surface.

First, let's find the radius of the neutron star (r). The diameter (D) is given as 25.0 km, so the radius ( r ) is half of the diameter:

[tex]\[ r = \frac{D}{2} = \frac{25.0 \ \text{km}}{2} = 12.5 \ \text{km} \][/tex]

Now, convert the radius to meters (1 km = 1000 m):

[tex]\[ r = 12.5 \ \text{km} \times 1000 \ \text{m/km} = 1.25 \times 10^4 \ \text{m} \][/tex]

Now, substitute the values into the gravitational force formula:

[tex]\[ F = \frac{{G \cdot M \cdot m}}{{r^2}} \][/tex]

[tex]\[ F = \frac{{(6.67 \times 10^{-11} \ \text{N} \cdot \text{m}^2/\text{kg}^2) \cdot (1.99 \times 10^{30} \ \text{kg}) \cdot (m)}}{{(1.25 \times 10^4 \ \text{m})^2}} \][/tex]

Now, set this force equal to the weight on Earth:

[tex]\[ \frac{{G \cdot M \cdot m}}{{r^2}} = m \cdot g \][/tex]

Solve for (m):

[tex]\[ m = \frac{{r^2 \cdot g}}{{G \cdot M}} \][/tex]

Plug in the values:

[tex]\[ m = \frac{{(1.25 \times 10^4 \ \text{m})^2 \cdot (9.81 \ \text{m/s}^2)}}{{6.67 \times 10^{-11} \ \text{N} \cdot \text{m}^2/\text{kg}^2 \cdot 1.99 \times 10^{30} \ \text{kg}}} \][/tex]

Now, calculate \( m \), and then use \( W = m \cdot g \) to find the weight on the surface of the neutron star.

Let's calculate the mass (m) first:

[tex]\[ m = \frac{{(1.25 \times 10^4 \ \text{m})^2 \cdot (9.81 \ \text{m/s}^2)}}{{6.67 \times 10^{-11} \ \text{N} \cdot \text{m}^2/\text{kg}^2 \cdot 1.99 \times 10^{30} \ \text{kg}}} \][/tex]

[tex]\[ m = \frac{{1.5625 \times 10^8 \ \text{m}^2 \cdot 9.81 \ \text{m/s}^2}}{{1.32733 \times 10^{20} \ \text{N}}} \][/tex]

[tex]\[ m \approx \frac{{1.5331875 \times 10^9 \ \text{m}^2}}{{1.32733 \times 10^{20} \ \text{N}}} \][/tex]

[tex]\[ m \approx 1.155 \times 10^{-11} \ \text{kg} \][/tex]

Now, calculate the weight (W) on the surface of the neutron star:

[tex]\[ W = m \cdot g \][/tex]

[tex]\[ W \approx (1.155 \times 10^{-11} \ \text{kg}) \cdot (9.81 \ \text{m/s}^2) \][/tex]

[tex]\[ W \approx 1.133 \times 10^{-10} \ \text{N} \][/tex]

So, if you weigh 660 N on Earth, you would weigh approximately[tex]\(1.133 \times 10^{-10}\)[/tex]N on the surface of a neutron star with the same mass as our sun and a diameter of 25.0 km.

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Answer:

The average acceleration is 8.06 m/s².

Explanation:

It is given that,

Initial speed of the jet, u = 120 mph = 176 ft/s

Final velocity of the jet, v = 30 mph = 44 ft/s

Distance, d = 1800 ft

We need to find the average acceleration required of the airplane during braking. It can be calculated using third law of motion as :

[tex]v^2-u^2=2ad[/tex]

a = acceleration

[tex]a=\dfrac{v^2-u^2}{2d}[/tex]

[tex]a=\dfrac{(44\ ft/s)^2-(176\ ft/s)^2}{2\times 1800\ ft}[/tex]

[tex]a=-8.06\ ft/s^2[/tex]

So, the average acceleration required of the airplane during braking is -8.06 ft/s². Hence, this is the required solution.

Answer:

Maximum height, h = 11.32 meters

Explanation:

It is given that,

The baseball is thrown directly upward at time, t = 0

Initial speed of the baseball, u = 14.9 m/s

Ignoring the resistance in this case and using a = g = 9.8 m/s²

We have to find the maximum height the ball reaches above where it leaves your hand. Let the maximum height is h. Using third equation of motion as :

[tex]v^2-u^2=2ah[/tex]

At maximum height, v = 0

and a = -g = -9.8 m/s²

[tex]h=\dfrac{v^2-u^2}{2a}[/tex]

[tex]h=\dfrac{0-(14.9\ m/s)^2}{2\times -9.8\ m/s^2}[/tex]

h = 11.32 meters

Hence, the maximum height of the baseball is 11.32 meters.

Answer:

The coefficient of friction is 0.56

Explanation:

It is given that,

Mass of the automobile, m = 1400 kg

Speed of the automobile, v = 23 m/s

Radius of the track, r = 95 m

The automobile is moving in a circular track. The centripetal force is given by :

[tex]F_c=\dfrac{mv^2}{r}[/tex]............(1)

Frictional force is given by :

[tex]F_f=\mu mg[/tex]...................(2)

[tex]\mu[/tex] = coefficient of friction

g = acceleration due to gravity

From equation (1) and (2), we get :

[tex]\dfrac{mv^2}{r}=\mu mg[/tex]

[tex]\mu=\dfrac{v^2}{rg}[/tex]

[tex]\mu=\dfrac{(23\ m/s)^2}{95\ m\times 9.8\ m/s^2}[/tex]

[tex]\mu=0.56[/tex]

So, the coefficient of friction is 0.56. Hence, this is the required solution.

To find the coefficient of friction, calculate the maximum speed at which the car can negotiate the curve without slipping using the formula v = √(μrg). Plug in the values and solve for μ to get the coefficient of friction. The approximate value is 0.24.

To find the coefficient of friction, we will first calculate the maximum speed at which the car can negotiate the curve without slipping. The maximum speed is given by the formula:

v = √(μrg)

Where v is the maximum speed, μ is the coefficient of friction, r is the radius of the curve, and g is the acceleration due to gravity.

Plugging in the values, we get:

23 = √(μ * 1400 * 9.8 * 95)

Squaring both sides of the equation, we have:

529 = μ * 1400 * 9.8 * 95

Simplifying the equation, we find:

μ = 529 / (1400 * 9.8 * 95)

Calculating the value, we get:

μ ≈ 0.24

Therefore, the coefficient of friction is approximately 0.24.

The radius of the circle around which samples travel in the mentioned ultracentrifuge is approximately 4.18 cm. This was calculated using the centripetal acceleration and the rotational speed of the ultracentrifuge, demonstrating how principles of physics apply to scientific and medical practices.

An ultracentrifuge utilizes very high speeds and centripetal acceleration to separate minute biological components in a sample. Given that the ultracentrifuge mentioned spins at 120,000 rotations per minute (rpm) and the centripetal acceleration recorded is 6.6×10⁶ m/s², we need to determine the radius of the circle around which these samples travel. This can be formulated using the physics concept of uniform circular motion.

To solve this, we will use the formula for centripetal acceleration, which is a = ω²r, where 'a' is centripetal acceleration, 'ω' is angular velocity, and 'r' is the radius of the circle. First, we need to convert the rotational speed from rpm to radians per second: 1 rpm is equal to 2π rad/60 s, so 120,000 rpm equals 2π * 120,000 / 60 = 12,566 rad/s. Then, we substitute these values into the formula and solve for the radius, giving us r = a/ω² ≈ 0.0418 meters or 4.18 cm.

The samples would therefore travel around a circular path with a radius of approximately 4.18 cm. This is a great example of how the principles of physics are used in scientific and medical practices, like the operation of an ultracentrifuge.

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The radius of the circle around which the samples travel in the ultracentrifuge is approximately 13.2 millimeters.

Calculating the Radius of Circular Motion in an Ultracentrifuge

To determine the radius of the circle around which the samples travel in an ultracentrifuge, use the formula for centripetal acceleration: a = ω²r, where a is the centripetal acceleration, ω is the angular velocity in radians per second, and r is the radius.

First, convert the given rotational speed from revolutions per minute (rpm) to radians per second:

Given: 120,000 rpm

Convert rpm to revolutions per second: 120,000 rpm ÷ 60 seconds/min = 2,000 revolutions per second

Convert revolutions per second to radians per second (since there are 2π radians in one revolution): 2,000 rev/s × 2π radians/rev = 4,000π radians/second

Now that we have the angular velocity, we can use the centripetal acceleration formula. Rearrange the formula to solve for radius r:

r = a / ω²

Substitute the given values (where a = 6.6×106 m/s² and ω = 4,000π rad/s):

r = 6.6×106 m/s² / (4,000π rad/s)²

Calculate the radius:

Square the angular velocity: (4,000π rad/s)² = 16,000,000π² rad²/s²

Divide the acceleration by the squared angular velocity: r = 6.6×[tex]10^6[/tex] m/s² / 16,000,000π² rad²/s² = 0.131 / π² m

Simplify using π ≈ 3.14159: r ≈ 0.131 / (3.14159)² ≈ 0.0132 m or 13.2 mm

Thus, the radius of the circle around which the samples travel in the ultracentrifuge is approximately 13.2 millimeters.

Answer:

Angle of banking be 15.64 degrees.

Explanation:

It is given that,

Radius of banked road comer, r = 146 m

Velocity of vehicle, v = 20 m/s

Banking angle is defined as the angle at which a vehicle is in inclined position. Angle of banking is given by :

[tex]tan\ \theta=\dfrac{v^2}{rg}[/tex]  , g = acceleration due to gravity

[tex]tan\ \theta=\dfrac{(20\ m/s)^2}{146\ m\times 9.8\ m/s^2}[/tex]

[tex]tan\ \theta=0.28[/tex]

[tex]\theta=tan^{-1}(0.28)[/tex]

[tex]\theta=15.64\ degrees[/tex]

Where, [tex]\theta[/tex] = angle of banking

Hence, this is the required solution.

Answer:

6.6 N

Explanation:

Let's take the direction of the force of 4.0 N as positive x-direction. This means that the force of 3.0 N is at 40 degrees above it. So the components of the two forces along the x- and y-directions are:

[tex]F_{1x} = 4.0 N\\F_{1y} = 0[/tex]

[tex]F_{2x} = 3.0 N cos 40^{\circ}=2.3 N\\F_{2y} = 3.0 N sin 40^{\circ} = 1.9 N[/tex]

So the resultant has components

[tex]F_x = F_{1x}+F_{2x}=4.0 N +2.3 N = 6.3 N\\F_y = F_{1y} + F_{2y} = 0 + 1.9 N = 1.9 N[/tex]

So the magnitude of the resultant is

[tex]F=\sqrt{F_x^2 +F_y^2}=\sqrt{(6.3)^2+(1.9)^2}=6.6 N[/tex]

And in order for the body to be balanced, the third force must be equal and opposite (in direction) to this force: so, the magnitude of the third force must be 6.6 N.

The magnitude of the resultant force when a 40 N force acts in the x-direction and a 30 N force acts in the y-direction is calculated using the Pythagorean theorem. Since these forces are perpendicular, the resultant force is the hypotenuse of a right triangle formed by the two forces, which equals 50 N.

Given a force of 40 N in the positive x-direction and a force of 30 N in the positive y-direction, these forces are perpendicular to each other. Hence, the resultant force is the hypotenuse of a right-angled triangle where the sides are the two forces.

To find the magnitude of the resultant force (R), we use the formula:

R = sqrt((40 N)^2 + (30 N)^2)

Calculate the R value (with calculator):

R = sqrt((1600) + (900))

R = sqrt(2500 N^2)

R = 50 N

Therefore, the magnitude of the resultant force is 50 N.

Answer:

At 11.5 m

Explanation:

The power per unit area corresponds to the intensity, which is given by

[tex]I=\frac{P}{4\pi r^2}[/tex]

where

P is the power

[tex]4\pi r^2[/tex] is the area irradiated at a distance r from the source (it corresponds to the surface area of a sphere of radius r)

Here we want the intensity of the two light bulbs to be the same, so

[tex]I_1 = I_2\\\frac{P_1}{4 \pi r_1^2}=\frac{P_2}{4\pi r_2^2}[/tex]

where we have

P1 = 100 W is the power of the first light bulb

P2 = 75 W is the power of the second light bulb

r2 = 10 m is the distance from the second light bulb

Solving for r1, we find

[tex]r_1 = r_2 \sqrt{\frac{P_1}{P_2}}= (10 m) \sqrt{\frac{100 W}{75 W}} = 11.5 m[/tex]

The distance at which a 100 Watt lightbulb delivers the same power per unit surface area as a 75 Watt lightbulb at 10m is approximately 14.6 meters.

The question is asking for the distance at which a 100 Watt lightbulb delivers the same power per unit surface area as a 75 Watt lightbulb at 10 m away. The intensity (power per unit area) of the light from a bulb diminishes with the square of the distance from the source. This is because the power (in Watts) is distributed over the surface area of a sphere, and the surface area of a sphere increases with the square of the radius (distance), following the formula 4πr².

First, we calculate the intensity I75 of the 75 Watt bulb at 10 m distance. That gives us: I75 = P75/4πr² = 75W/4π(10m)² ≈ 0.60 W/m².

Now, we want to find the distance r100 at which the 100 Watt bulb would deliver this same intensity. Solving the similar formula for r100 gives us: r100 = sqrt(P100/4πI75) = sqrt(100W/4π(0.60 W/m²)) ≈ 14.6 m.

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Answer:

4.42 x 10⁷ W/m²

Explanation:

A = energy absorbed = 500 J

η = efficiency = 0.90

E = Total energy

Total energy is given as

E = A/η

E = 500/0.90

E = 555.55 J

t = time = 4.00 s

Power of the beam is given as

P = E /t

P = 555.55/4.00

P = 138.88 Watt

d = diameter of the circular spot = 2.00 mm = 2 x 10⁻³ m

Area of the circular spot is given as

A = (0.25) πd²

A = (0.25) (3.14) (2 x 10⁻³)²

A = 3.14 x 10⁻⁶ m²

Intensity of the beam is given as

I = P /A

I = 138.88 / (3.14 x 10⁻⁶)

I = 4.42 x 10⁷ W/m²

The intensity of a laser beam : 4.42.10⁷ W/m²

The energy transferred by waves per unit area per unit time is called wave intensity  

Because energy per unit time is Power, the intensity of the wave is equal to Power divided by area  

[tex]\rm P=\dfrac{W}{t}[/tex]

P = power, watt

W = energy, J  

t = time, s  

For waves that spread in all directions, the intensity at the distance R from the source can be formulated  

[tex]\rm I=\dfrac{Power}{Area}=\dfrac{P}{\pi .R^2}[/tex]

From the equation above shows the intensity of the wave is inversely proportional to the square of the distance from the source.  

[tex]\rm I\approx \dfrac{1}{R^2}[/tex]

The farther the wave spreads, the smaller the intensity  

Cancerous tissue area:

[tex]\rm A=\dfrac{1}{4}\pi d^2\\\\A=\dfrac{1}{4}\pi(2.10^{-3})^2\Rightarrow d=2~mm=2.10^{-3}\:m\\\\A=\frac{1}{4}\pi 4.10^{-6}\\\\A=\pi .10^{-6}\\\\A=3.14.10^{-6}[/tex]

So that the intensity

[tex]\rm 0.9I(only~90\%~absorbed)=\dfrac{W}{A.t}\\\\0.9I=\dfrac{500}{3.14.10^{-6}.4}\\\\I=4.42.10^7~\dfrac{W}{m^2}[/tex]

electric field

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magnetism

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An electric device delivers a current of 5 a to a device.

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Answer:

124.3 J/m^3

Explanation:

The energy density between the plates of a parallel-plate capacitor is given by

[tex]u=\frac{1}{2}\epsilon_0 E^2[/tex]

where

ε0 = 8.85 × 10-12 C2/N · m2 is the vacuum permittivity

E is the electric field strength

In this problem,

E = 5.3 × 106 V/m

So the energy density is

[tex]u=\frac{1}{2}(8.85\cdot 10^{-12}F/m) (5.3\cdot 10^6 V/m)^2=124.3 J/m^3[/tex]

The energy density between the plates of a capacitor can be calculated using the formula U = 0.5 * ε0 * E². By substituting the given values for the electric field (E) and the permittivity of free space (ε0) into the formula, we can ascertain the energy density.

To find the energy density between the plates of the capacitor we can use the formula U = 0.5 * ε0 * E².

First, we need to find the electric field (E) which is given to us as 5.3 × 10^6 V/m. The permittivity of free space (ε0) is 8.85 × 10^-12 C^2/N * m^2.

Then we substitute the values into the formula to calculate the energy density:

U = 0.5 * ε0 * E² = 0.5 * 8.85 × 10^-12 C^2/N * m^2 * (5.3 × 10^6 V/m)²

So, the energy density between the plates of the capacitor is calculated by using the given values of the electric field and permittivity of free space.

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Answer:

7.54 m/s²

Explanation:

uₓ(t) = (0.980 m/s³) t²

Acceleration is the derivative of velocity with respect to time.

aₓ(t) = 2 (0.980 m/s³) t

aₓ(t) = (1.96 m/s³) t

When uₓ = 14.5 m/s, the time is:

14.5 m/s = (0.980 m/s³) t²

t = 3.85 s

Plugging into acceleration equation:

aₓ = (1.96 m/s³) (3.85 s)

aₓ = 7.54 m/s²

This question is dealing with velocity, acceleration and time of motion.

Acceleration is; aₓ = 7.54 m/s²

We are told that the eastward component of the car's velocity is;

uₓ(t) = (0.980 m/s³) t²

Now, from calculus differentiation in maths, we know that with respect to time, the derivative of velocity is equal to the acceleration.

Thus;

aₓ(t) = du/dt = 2t(0.980 m/s³)  

aₓ(t) = 1.96t m/s³

We w ant to find the acceleration of the car when velocity is; uₓ = 14.5 m/s. Let us find the time first and then plug the value into the acceleration equation.

Thus;

14.5 m/s = (0.980 m/s³) t²

14.5/0.98 = t²

t = 3.85 s

Putting 3.85 for t in the acceleration equation to get;

aₓ = (1.96 m/s³) (3.85 s)

aₓ = 7.54 m/s²

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Explanation:

It is given that,

Radius of circular particle accelerator, r = 1 m

The distance covered by the particle is equal to the circumference of the circular path, d = 2πr

d = 2π × 1 m

(a) The speed of satellite is given by total distance divided by total time taken as :

[tex]speed=\dfrac{distance}{time}[/tex]

Let t is the period of the particle.

[tex]t=\dfrac{d}{s}[/tex]

d = distance covered

s = speed of particle

It is given that the charged particle is moving nearly with the speed of light

[tex]t=\dfrac{d}{c}[/tex]

[tex]t=\dfrac{2\pi\times 1\ m}{3\times 10^8\ m/s}[/tex]

[tex]t=2.09\times 10^{-8}\ s[/tex]

(b) On the circular path, the centripetal acceleration is given by :

[tex]a=\dfrac{c^2}{r}[/tex]

[tex]a=\dfrac{(3\times 10^8\ m/s)^2}{1\ m}[/tex]

[tex]a=9\times 10^{16}\ m/s^2[/tex]

Hence, this is the required solution.

Answer:

The weights are 1 kg, 3kg, 9kg and 27kg.

Explanation:

The weights are 1 kg, 3kg, 9kg and 27kg.

1+3+9+27= 40

27+9+3= 39

27+9+3-1=38

27+9+1=37

27+9=36

27+9-1=35

27+9+1-3=34

27+9-3=33

27+9-3-1=32

27+3+1=31

27+3=30

27+3-1=29

27+1=28

27

27-1=26

27+1-3=25

27-3=24

27-3-1=23

27+3+1-9=22

27+3-9=21

27+3-9-1=20

Like this all the weights from 1 to 40 kg can be made using 1,3,9 and 27 kg.

Final answer:

Using the kinetic energy formula, the 500-kg object moving at 40 m/s has a kinetic energy of 400,000 J, while the 1000-kg object moving at 20 m/s has a kinetic energy of 200,000 J. Therefore, the 500-kg object has larger kinetic energy.

Explanation:

To determine which object has larger kinetic energy, we can use the formula Ek = 1/2 m v2, where Ek is the kinetic energy, m is the mass of the object, and v is the velocity of the object. Plugging in the values for both objects:

Comparing the results, the 500-kg object has larger kinetic energy than the 1000-kg object. Therefore, the correct answer is:

a. The 500-kg object

Answer:

Part a)

[tex]R_1 = 0.072 m[/tex]

Part b)

[tex]R_2 = 0.036 m[/tex]

Part c)

d = 0.072 m

Explanation:

Part a)

As we know that the radius of the charge particle in constant magnetic field is given by

[tex]R = \frac{mv}{qB}[/tex]

now for single ionized we have

[tex]R_1 = \frac{(4.39\times 10^{-27})(7.92 \times 10^5)}{(1.6 \times 10^{-19})(0.301)}[/tex]

[tex]R_1 = 0.072 m[/tex]

Part b)

Similarly for doubly ionized ion we will have the same equation

[tex]R = \frac{mv}{qB}[/tex]

[tex]R_2 = \frac{(4.39\times 10^{-27})(7.92 \times 10^5)}{(3.2 \times 10^{-19})(0.301)}[/tex]

[tex]R_2 = 0.036 m[/tex]

Part c)

The distance between the two particles are half of the loop will be given as

[tex]d = 2(R_1 - R_2)[/tex]

[tex]d = 2(0.072 - 0.036)[/tex]

[tex]d = 0.072 m[/tex]